Q: $252$ can be expressed as a product of primes as

Concept & Strategy: Expressing a number "as a product of primes" simply means calculating its complete prime factorization. The most reliable method is creating a factor tree or using the ladder division method starting from the smallest prime. Step-by-Step Solution: • Step 1: 252 is even, so divide by 2. 252 \div 2 = 126 • Step 2: 126 is also even, divide by 2 again. 126 \div 2 = 63 • Step 3: 63 is not even. The sum of digits (6+3=9) is divisible by 3. Divide by 3. 63 \div 3 = 21 • Step 4: 21 is notoriously divisible by 3. 21 \div 3 = 7 • Step 5: 7 is a prime number. Divide by 7. 7 \div 7 = 1 • Calculation / Deduction: Writing out all the divisors gathered in our steps: 252 = 2 × 2 × 3 × 3 × 7 Exam Strategy & Shortcut: Direct Option Testing: Since you are given the expressions, calculate their products quickly. • (a) 4 × 9 × 7 = 36 × 7 = 252. (Match found!) • (b) 8 × 21 = 168. (Incorrect) • (c) 81 × 7 = 567. (Incorrect) • (d) 54 × 7 = 378. (Incorrect) This method is often faster than performing the division steps yourself. Common Pitfall: Dividing by composite numbers (like 4 or 9) initially to save time, and then forgetting to break those composite numbers down into primes at the final step. Always ensure every number in the final list is strictly prime. Final Answer: Therefore, the correct answer is 2 × 2 × 3 × 3 × 7.

Q: A number $n$ is said to be perfect if the sum of all its divisors (excluding $n$ itself) is equal to $n$. An example of perfect number is

Concept & Strategy: A "Perfect Number" is a positive integer that is equal to the sum of its proper divisors. Proper divisors are all the positive integers that divide the number evenly, excluding the number itself. The strategy here is straightforward: extract the proper divisors for each option and calculate their sum to see which one equals the original number. Step-by-Step Solution: We will test each option to find the perfect number: • Test Option (a) 6: Divisors of 6 (excluding 6): 1, 2, 3 Sum of divisors = 1 + 2 + 3 = 6 Since 6 = 6, this is a perfect number! • Test Option (b) 9: Divisors of 9 (excluding 9): 1, 3 Sum of divisors = 1 + 3 = 4 Since 4 \neq 9, it is not a perfect number. • Test Option (c) 15: Divisors of 15 (excluding 15): 1, 3, 5 Sum of divisors = 1 + 3 + 5 = 9 Since 9 \neq 15, it is not a perfect number. • Test Option (d) 21: Divisors of 21 (excluding 21): 1, 3, 7 Sum of divisors = 1 + 3 + 7 = 11 Since 11 \neq 21, it is not a perfect number. Exam Strategy & Shortcut: Memorization Benchmark: Perfect numbers are rare in early mathematics. The first few perfect numbers are mathematically famous: 6, 28, 496, and 8128. Memorizing at least the first two (6 and 28) is highly recommended for aptitude exams as they frequently appear as trivia or direct questions, allowing you to bypass calculation entirely. Common Pitfall: A common error is forgetting to exclude the number itself (n) when calculating the sum of divisors. If you include the number itself, the sum for a perfect number will always equal 2n instead of n. Final Answer: Therefore, the correct answer is 6.

Q: $\frac{1095}{1168}$ when expressed in simplest form is

Concept & Strategy: To reduce a fraction to its simplest form, we need to find the Highest Common Factor (HCF) of its numerator and denominator, and then divide both by this HCF. Alternatively, we can use the difference method: the HCF of two numbers divides their difference. This helps quickly narrow down potential factors without doing long division. Step-by-Step Solution: • Given: The fraction is (1095/1168). • Calculation / Deduction: 1. Let's find the difference between the denominator and the numerator: 1168 - 1095 = 73 2. The HCF must be a factor of 73. Since 73 is a prime number, the HCF is highly likely to be 73. 3. Test if both numbers are divisible by 73: 1095 \div 73 = 15 1168 \div 73 = 16 4. Since both divide perfectly, we reduce the fraction by dividing the top and bottom by 73: (1095 \div 73/1168 \div 73) = (15/16) Exam Strategy & Shortcut: Unit Digit Estimation: Look at the options. The fraction ends in 5 on top and 8 on the bottom. We need a multiplier k such that k × Numerator Option ends in 5, and k × Denominator Option ends in 8. Let's test option (b) (15/16). If 15 × k ends in 5, k must be odd (like 3, 5, 7, \dots). If 16 × k ends in 8, k could be 3 (16 × 3 = 48) or 8 (16 × 8 = 128). Let's test k ending in 3: 15 × 3 = 45 (ends in 5), 16 × 3 = 48 (ends in 8). Scaling this up, let k = 73. 15 × 73 = 1095. 16 × 73 = 1168. The match is instantaneous. Common Pitfall: Trying to systematically divide 1095 and 1168 by small primes (2, 3, 5, 7, 11\dots). Since their only common factor is the large prime 73, you will waste immense time and likely assume the fraction cannot be reduced. Always use the difference method for large numbers. Final Answer: Therefore, the correct answer is (15/16).

Q: Reduce $\frac{128352}{238368}$ to its lowest terms.

Concept & Strategy: Reducing massive fractions like (128352/238368) using standard division or finding the HCF is too time-consuming for an aptitude exam. Instead, we reverse-engineer the problem using the given multiple-choice options. We assume one of the options is correct, which means its denominator multiplied by some factor k equals 238368, and its numerator multiplied by the same k equals 128352. Step-by-Step Solution: • Given: The fraction (128352/238368). • Calculation / Deduction: 1. First, quickly estimate the value of the fraction to eliminate wild options. (128/238) is roughly (12/24) or (1/2) (0.5). 2. Check the options based on this estimate: (3/4) = 0.75 (Too big) (5/13) \approx 0.38 (Too small) (7/13) \approx 0.53 (Very close) (9/13) \approx 0.69 (Too big) 3. Option (c) is the only logical candidate. Let's verify it strictly. 4. If (7/13) is the exact ratio, then 238368 \div 13 should give us our exact multiplier k. 238368 \div 13 = 18336 5. Now multiply this multiplier k by the numerator 7: 18336 × 7 = 128352 6. This perfectly matches the original numerator, confirming our ratio. Exam Strategy & Shortcut: Cross-Multiplication Unit Digit Check: If (128352/238368) = (7/13), then 128352 × 13 = 238368 × 7. Check just the unit digits: Left side: 2 × 3 = 6. Right side: 8 × 7 = 56 (ends in 6). The unit digits match perfectly. Doing this mentally takes 5 seconds and solidly confirms your estimation. Common Pitfall: Students often attempt long division to find the HCF of 128352 and 238368. For six-digit numbers, this is a trap designed to drain your exam time. Never calculate the HCF directly if simple fractional options are provided. Final Answer: Therefore, the correct answer is (7/13).

Q: The simplest reduction to the lowest terms of $\frac{116,690,151}{427,863,887}$ is

Concept & Strategy: For extremely large, nine-digit fractions, the problem is testing your ability to use options and estimation, rather than your ability to calculate rigorous prime factorizations. We treat the multiple-choice options as ratios and check which denominator perfectly scales up to the given large denominator, and verify if the numerator scales up by that exact same multiple. Step-by-Step Solution: • Given: The fraction (116690151/427863887). • Calculation / Deduction: 1. Look at the leading digits to estimate the ratio: (116/427). 2. Notice that 116 × 3 = 348, and 116 × 4 = 464. The value is slightly more than (1/4) (0.25). 3. Evaluate the given options based on this rough estimate: (a) (3/11) \approx 0.27 (A strong candidate) (b) (7/11) \approx 0.63 (Incorrect) (c) (11/3) = 3.66 (Incorrect) 4. Let's test option (a) rigorously. If the denominator ratio is 11, then 1 "part" is: 427863887 \div 11 = 38896717 5. Now, multiply this 1-part value by the numerator ratio 3: 38896717 × 3 = 116690151 6. This matches the original numerator perfectly. Exam Strategy & Shortcut: Digit Sum / Divisibility Rule Check: Let's test divisibility by 11 on the denominator 427863887. Rule for 11: (Sum of odd-placed digits) - (Sum of even-placed digits). Odd places: 4+7+6+8+7 = 32. Even places: 2+8+3+8 = 21. Difference: 32 - 21 = 11. Since 11 is a multiple of 11, the denominator is divisible by 11, making options (a) and (b) highly likely. Combining this with our basic (1/4) estimation immediately isolates (3/11). Common Pitfall: Choosing "None of these" because the numbers look too intimidating to check. Examiners use terrifyingly large numbers specifically to reward candidates who stay calm and apply basic approximation and divisibility rules. Final Answer: Therefore, the correct answer is (3/11).

Q: The H.C.F. of $2^2 \times 3^3 \times 5^5$, $2^3 \times 3^2 \times 5^2 \times 7$ and $2^4 \times 3^4 \times 5 \times 7^2 \times 11$ is

Concept & Formula: To find the Highest Common Factor (HCF) of numbers already given in their prime factorized form, you do not need to expand or multiply them out. The Rule for HCF of Prime Expressions: Select only the prime bases that are common to all the given numbers, and raise each to its lowest power present across the expressions. Step-by-Step Solution: • Given: We have three numbers expressed as products of primes: Number 1: 2^2 × 3^3 × 5^5 Number 2: 2^3 × 3^2 × 5^2 × 7 Number 3: 2^4 × 3^4 × 5 × 7^2 × 11 • Calculation / Deduction: 1. Identify Common Bases: Look at the prime bases (2, 3, 5, 7, 11). Only the bases 2, 3, and 5 appear in all three expressions. (We ignore 7 and 11 because they are missing from the first expression). 2. Find the Lowest Power of 2: The powers of 2 are 2^2, 2^3, 2^4. The lowest is 2^2. 3. Find the Lowest Power of 3: The powers of 3 are 3^3, 3^2, 3^4. The lowest is 3^2. 4. Find the Lowest Power of 5: The powers of 5 are 5^5, 5^2, 5^1 (since 5 is 5^1). The lowest is 5^1 or just 5. 5. Multiply these lowest powers together to get the HCF: HCF = 2^2 × 3^2 × 5 Exam Strategy & Shortcut: Rapid Elimination: You can solve this by scanning the options instantly. Since the first expression lacks 7 and 11, any option containing 7 or 11 is immediately incorrect. This quickly eliminates options (b) and (d). Next, look at the base 5. The third expression only has 5^1. Therefore, the HCF cannot have a power of 5 greater than 1. This eliminates option (c) which contains 5^5. Option (a) is the only survivor. Common Pitfall: Confusing the rules for HCF and LCM. Students often mistakenly select the *highest* powers or include all distinct prime factors (which calculates the LCM, not HCF). Remember: HCF equals Common bases only, Lowest powers. Final Answer: Therefore, the correct answer is 2^2 × 3^2 × 5.

Q: The H.C.F. of $2^4 \times 3^2 \times 5^3 \times 7$, $2^3 \times 3^3 \times 5^2 \times 7^2$ and $3 \times 5 \times 7 \times 11$ is

Concept & Strategy: To find the Highest Common Factor (HCF) of numbers provided in their prime factorized form, you only need to look at the prime bases. The rule is: Identify the prime bases that are common to all expressions, and take the lowest power of each common base present. Multiply these together to get the HCF. Step-by-Step Solution: • Given: Expression 1: 2^4 × 3^2 × 5^3 × 7^1 Expression 2: 2^3 × 3^3 × 5^2 × 7^2 Expression 3: 3^1 × 5^1 × 7^1 × 11^1 • Calculation / Deduction: 1. Identify Common Bases: Look at the prime bases across all three expressions. The bases 3, 5, and 7 appear in every expression. The base 2 is missing from the third expression, and 11 is missing from the first two. 2. Lowest Power of 3: Comparing 3^2, 3^3, and 3^1, the lowest is 3^1 (or simply 3). 3. Lowest Power of 5: Comparing 5^3, 5^2, and 5^1, the lowest is 5^1 (or simply 5). 4. Lowest Power of 7: Comparing 7^1, 7^2, and 7^1, the lowest is 7^1 (or simply 7). 5. Calculate Final HCF: Multiply these common lowest powers together: HCF = 3 × 5 × 7 HCF = 15 × 7 = 105 Exam Strategy & Shortcut: Rapid Elimination: You can spot the answer in under 10 seconds. The third expression is just 3 × 5 × 7 × 11. Since 2 is missing, the HCF cannot be even. Options (c) 2310 and (d) 27720 are even numbers, so eliminate them instantly. Furthermore, 11 is not present in the first expression, so 11 cannot be in the HCF. Option (b) 1155 is divisible by 11 (1+5 = 6, 1+5 = 6). This leaves only 105 as the correct choice. Common Pitfall: The most frequent mistake is getting confused with the LCM rule and picking the *highest* powers, or blindly multiplying all distinct prime factors together. Always remember: HCF means "Common" bases and "Lowest" powers. Final Answer: Therefore, the correct answer is 105.

Question 1

The traffic lights at three different road crossings change after every 48 sec., 72 sec. and 108 sec. respectively. If they all change simultaneously at 8:20:00 hours, then at what time will they again change simultaneously?

Accepted Answer

Concept & Strategy: When periodic events (like bells ringing, lights changing, or people running a track) occur at different intervals, they will all happen together again at a time interval equal to the Lowest Common Multiple (L.C.M.) of their individual periods. Simultaneous Interval = L.C.M.(T_1, T_2, T_3, \dots) Step-by-Step Solution: • Given: Intervals of change are 48 sec, 72 sec, and 108 sec. Initial simultaneous change was at 8:20:00. • Find the L.C.M. of 48, 72, and 108 using prime factorization: • 48 = 2^4 × 3 • 72 = 2^3 × 3^2 • 108 = 2^2 × 3^3 • Extract the highest powers of the prime factors: • Highest power of 2 is 2^4 = 16. • Highest power of 3 is 3^3 = 27. • Multiply them to get the L.C.M.: • L.C.M. = 16 × 27 = 432 seconds. • This means the lights will change together again after 432 seconds. • Convert 432 seconds into minutes and seconds: • 432 / 60 = 7 minutes with a remainder of 12 seconds. • So, 432 seconds = 7 minutes 12 seconds. • Add this time interval to the initial time (8:20:00): • 8 hours : 20 minutes : 00 seconds • + 0 hours : 7 minutes : 12 seconds • = 8:27:12 hours. Exam Strategy & Shortcut: Instead of heavy prime factorization, look for common large factors to speed up L.C.M. calculation. All numbers are multiples of 12. • 48 = 12 × 4 • 72 = 12 × 6 • 108 = 12 × 9 The L.C.M. of (4, 6, 9) is simply 36. Multiply the common factor back: 12 × 36 = 432 seconds. Then convert to minutes rapidly. Common Pitfall: A careless mistake in time addition. Students calculate 432 seconds correctly but mess up the base-60 conversion (e.g., assuming 100 seconds in a minute), leading to an incorrect final timestamp. Always meticulously divide by 60 for time. Final Answer: Therefore, the correct answer is 8:27:12 hrs.

Question 2

Seema, Meena and Reema begin to jog around a circular stadium and they complete their revolutions in 54 seconds, 42 seconds and 63 seconds respectively. After how much time will they come together at the starting point?

Accepted Answer

Concept & Strategy: When multiple entities move in a circular path starting at the same time and place, they will all meet again at the starting point after a time equal to the Lowest Common Multiple (L.C.M.) of the times taken by each to complete one full revolution. Meeting Time = L.C.M.(T_1, T_2, T_3, \dots) Step-by-Step Solution: • Given: Revolution times are 54 seconds, 42 seconds, and 63 seconds. • Calculate the L.C.M. of 54, 42, and 63. Let's use prime factorization: • 54 = 2 × 3^3 • 42 = 2 × 3 × 7 • 63 = 3^2 × 7 • Take the highest power of each prime factor present: • For 2: 2^1 = 2 • For 3: 3^3 = 27 • For 7: 7^1 = 7 • Multiply to find the L.C.M.: • L.C.M. = 2 × 27 × 7 = 54 × 7 = 378 seconds. • The three girls will meet again at the starting point exactly 378 seconds later. • Convert 378 seconds into minutes and seconds: • 378 / 60 gives a quotient of 6 (minutes) and a remainder of 18 (seconds). • So, 378 seconds = 6 minutes 18 seconds. Exam Strategy & Shortcut: Use the options backwards. The correct total seconds must be a multiple of 63 (which means it must be a multiple of 9, so its digit sum is 9). Convert the options to seconds: • Option A: (6 × 60) + 18 = 378 (Digit sum = 18, div by 9. Works). • Option B: (6 × 60) + 36 = 396 (Digit sum = 18. Does it divide by 7? 396 / 7 = 56.5. Fails). • Option C: (5 × 60) + 48 = 348 (Fails digit sum of 9). Option A is confirmed in seconds without calculating the full LCM from scratch. Common Pitfall: Students sometimes confuse "meeting at the starting point" with "meeting anywhere on the track." If the question asks when they meet *anywhere*, relative speeds are required. Because they specify meeting *at the starting point*, pure L.C.M. of individual lap times is the only correct method. Final Answer: Therefore, the correct answer is 6 min 18 sec.

Question 3

Arrange the fractions $\frac{17}{18}$, $\frac{31}{36}$, $\frac{43}{45}$, $\frac{59}{60}$ in the ascending order.

Accepted Answer

Concept & Strategy: To reliably compare and order multiple fractions, convert them into equivalent fractions with a common denominator. The common denominator will be the L.C.M. of the original denominators. Once the denominators are identical, simply sort the fractions based on the values of their numerators. Step-by-Step Solution: • Given: The fractions are (17/18), (31/36), (43/45), and (59/60). • Find the L.C.M. of the denominators (18, 36, 45, and 60). • L.C.M. of 18, 36, 45, 60 = 180. • Convert each fraction to an equivalent fraction with a denominator of 180: • (17/18) = (17 × 10/18 × 10) = (170/180) • (31/36) = (31 × 5/36 × 5) = (155/180) • (43/45) = (43 × 4/45 × 4) = (172/180) • (59/60) = (59 × 3/60 × 3) = (177/180) • Compare the new numerators: • 155 < 170 < 172 < 177 • Therefore, their corresponding equivalent fractions follow the same ascending order: • (155/180) < (170/180) < (172/180) < (177/180) • Substitute back the original fractions: • (31/36) < (17/18) < (43/45) < (59/60) Exam Strategy & Shortcut: Use the Numerator-Denominator Difference Trick. Look at the gap between the numerator and denominator for each fraction: • (17/18) (Gap = 1) • (31/36) (Gap = 5) • (43/45) (Gap = 2) • (59/60) (Gap = 1) For proper fractions where values are close to 1, the fraction with the largest gap is generally the smallest. The gap of 5 implies (31/36) is the smallest. The gap of 2 implies (43/45) is next. The gap of 1 is shared by two fractions, so compare them directly: (17/18) vs (59/60). Since 59/60 is closer to 60/60, it is larger. Result: (31/36) < (17/18) < (43/45) < (59/60). Common Pitfall: Attempting to convert all fractions into exact decimals by long division under time pressure is a major mistake. It takes far too much time and increases the risk of calculation errors. Stick to L.C.M. equivalence or gap observation methods. Final Answer: Therefore, the correct answer is (31/36) < (17/18) < (43/45) < (59/60).

Question 4

$252$ can be expressed as a product of primes as

Accepted Answer

Concept & Strategy: Expressing a number "as a product of primes" simply means calculating its complete prime factorization. The most reliable method is creating a factor tree or using the ladder division method starting from the smallest prime. Step-by-Step Solution: • Step 1: 252 is even, so divide by 2. 252 \div 2 = 126 • Step 2: 126 is also even, divide by 2 again. 126 \div 2 = 63 • Step 3: 63 is not even. The sum of digits (6+3=9) is divisible by 3. Divide by 3. 63 \div 3 = 21 • Step 4: 21 is notoriously divisible by 3. 21 \div 3 = 7 • Step 5: 7 is a prime number. Divide by 7. 7 \div 7 = 1 • Calculation / Deduction: Writing out all the divisors gathered in our steps: 252 = 2 × 2 × 3 × 3 × 7 Exam Strategy & Shortcut: Direct Option Testing: Since you are given the expressions, calculate their products quickly. • (a) 4 × 9 × 7 = 36 × 7 = 252. (Match found!) • (b) 8 × 21 = 168. (Incorrect) • (c) 81 × 7 = 567. (Incorrect) • (d) 54 × 7 = 378. (Incorrect) This method is often faster than performing the division steps yourself. Common Pitfall: Dividing by composite numbers (like 4 or 9) initially to save time, and then forgetting to break those composite numbers down into primes at the final step. Always ensure every number in the final list is strictly prime. Final Answer: Therefore, the correct answer is 2 × 2 × 3 × 3 × 7.

Question 5

A number $n$ is said to be perfect if the sum of all its divisors (excluding $n$ itself) is equal to $n$. An example of perfect number is

Accepted Answer

Concept & Strategy: A "Perfect Number" is a positive integer that is equal to the sum of its proper divisors. Proper divisors are all the positive integers that divide the number evenly, excluding the number itself. The strategy here is straightforward: extract the proper divisors for each option and calculate their sum to see which one equals the original number. Step-by-Step Solution: We will test each option to find the perfect number: • Test Option (a) 6: Divisors of 6 (excluding 6): 1, 2, 3 Sum of divisors = 1 + 2 + 3 = 6 Since 6 = 6, this is a perfect number! • Test Option (b) 9: Divisors of 9 (excluding 9): 1, 3 Sum of divisors = 1 + 3 = 4 Since 4 
eq 9, it is not a perfect number. • Test Option (c) 15: Divisors of 15 (excluding 15): 1, 3, 5 Sum of divisors = 1 + 3 + 5 = 9 Since 9 
eq 15, it is not a perfect number. • Test Option (d) 21: Divisors of 21 (excluding 21): 1, 3, 7 Sum of divisors = 1 + 3 + 7 = 11 Since 11 
eq 21, it is not a perfect number. Exam Strategy & Shortcut: Memorization Benchmark: Perfect numbers are rare in early mathematics. The first few perfect numbers are mathematically famous: 6, 28, 496, and 8128. Memorizing at least the first two (6 and 28) is highly recommended for aptitude exams as they frequently appear as trivia or direct questions, allowing you to bypass calculation entirely. Common Pitfall: A common error is forgetting to exclude the number itself (n) when calculating the sum of divisors. If you include the number itself, the sum for a perfect number will always equal 2n instead of n. Final Answer: Therefore, the correct answer is 6.

Question 6

$\frac{1095}{1168}$ when expressed in simplest form is

Accepted Answer

Concept & Strategy: To reduce a fraction to its simplest form, we need to find the Highest Common Factor (HCF) of its numerator and denominator, and then divide both by this HCF. Alternatively, we can use the difference method: the HCF of two numbers divides their difference. This helps quickly narrow down potential factors without doing long division. Step-by-Step Solution: • Given: The fraction is (1095/1168). • Calculation / Deduction: 1. Let's find the difference between the denominator and the numerator: 1168 - 1095 = 73 2. The HCF must be a factor of 73. Since 73 is a prime number, the HCF is highly likely to be 73. 3. Test if both numbers are divisible by 73: 1095 \div 73 = 15 1168 \div 73 = 16 4. Since both divide perfectly, we reduce the fraction by dividing the top and bottom by 73: (1095 \div 73/1168 \div 73) = (15/16) Exam Strategy & Shortcut: Unit Digit Estimation: Look at the options. The fraction ends in 5 on top and 8 on the bottom. We need a multiplier k such that k × Numerator Option ends in 5, and k × Denominator Option ends in 8. Let's test option (b) (15/16). If 15 × k ends in 5, k must be odd (like 3, 5, 7, \dots). If 16 × k ends in 8, k could be 3 (16 × 3 = 48) or 8 (16 × 8 = 128). Let's test k ending in 3: 15 × 3 = 45 (ends in 5), 16 × 3 = 48 (ends in 8). Scaling this up, let k = 73. 15 × 73 = 1095. 16 × 73 = 1168. The match is instantaneous. Common Pitfall: Trying to systematically divide 1095 and 1168 by small primes (2, 3, 5, 7, 11\dots). Since their only common factor is the large prime 73, you will waste immense time and likely assume the fraction cannot be reduced. Always use the difference method for large numbers. Final Answer: Therefore, the correct answer is (15/16).

Question 7

Reduce $\frac{128352}{238368}$ to its lowest terms.

Accepted Answer

Concept & Strategy: Reducing massive fractions like (128352/238368) using standard division or finding the HCF is too time-consuming for an aptitude exam. Instead, we reverse-engineer the problem using the given multiple-choice options. We assume one of the options is correct, which means its denominator multiplied by some factor k equals 238368, and its numerator multiplied by the same k equals 128352. Step-by-Step Solution: • Given: The fraction (128352/238368). • Calculation / Deduction: 1. First, quickly estimate the value of the fraction to eliminate wild options. (128/238) is roughly (12/24) or (1/2) (0.5). 2. Check the options based on this estimate: (3/4) = 0.75 (Too big) (5/13) \approx 0.38 (Too small) (7/13) \approx 0.53 (Very close) (9/13) \approx 0.69 (Too big) 3. Option (c) is the only logical candidate. Let's verify it strictly. 4. If (7/13) is the exact ratio, then 238368 \div 13 should give us our exact multiplier k. 238368 \div 13 = 18336 5. Now multiply this multiplier k by the numerator 7: 18336 × 7 = 128352 6. This perfectly matches the original numerator, confirming our ratio. Exam Strategy & Shortcut: Cross-Multiplication Unit Digit Check: If (128352/238368) = (7/13), then 128352 × 13 = 238368 × 7. Check just the unit digits: Left side: 2 × 3 = 6. Right side: 8 × 7 = 56 (ends in 6). The unit digits match perfectly. Doing this mentally takes 5 seconds and solidly confirms your estimation. Common Pitfall: Students often attempt long division to find the HCF of 128352 and 238368. For six-digit numbers, this is a trap designed to drain your exam time. Never calculate the HCF directly if simple fractional options are provided. Final Answer: Therefore, the correct answer is (7/13).

Question 8

The simplest reduction to the lowest terms of $\frac{116,690,151}{427,863,887}$ is

Accepted Answer

Concept & Strategy: For extremely large, nine-digit fractions, the problem is testing your ability to use options and estimation, rather than your ability to calculate rigorous prime factorizations. We treat the multiple-choice options as ratios and check which denominator perfectly scales up to the given large denominator, and verify if the numerator scales up by that exact same multiple. Step-by-Step Solution: • Given: The fraction (116690151/427863887). • Calculation / Deduction: 1. Look at the leading digits to estimate the ratio: (116/427). 2. Notice that 116 × 3 = 348, and 116 × 4 = 464. The value is slightly more than (1/4) (0.25). 3. Evaluate the given options based on this rough estimate: (a) (3/11) \approx 0.27 (A strong candidate) (b) (7/11) \approx 0.63 (Incorrect) (c) (11/3) = 3.66 (Incorrect) 4. Let's test option (a) rigorously. If the denominator ratio is 11, then 1 "part" is: 427863887 \div 11 = 38896717 5. Now, multiply this 1-part value by the numerator ratio 3: 38896717 × 3 = 116690151 6. This matches the original numerator perfectly. Exam Strategy & Shortcut: Digit Sum / Divisibility Rule Check: Let's test divisibility by 11 on the denominator 427863887. Rule for 11: (Sum of odd-placed digits) - (Sum of even-placed digits). Odd places: 4+7+6+8+7 = 32. Even places: 2+8+3+8 = 21. Difference: 32 - 21 = 11. Since 11 is a multiple of 11, the denominator is divisible by 11, making options (a) and (b) highly likely. Combining this with our basic (1/4) estimation immediately isolates (3/11). Common Pitfall: Choosing "None of these" because the numbers look too intimidating to check. Examiners use terrifyingly large numbers specifically to reward candidates who stay calm and apply basic approximation and divisibility rules. Final Answer: Therefore, the correct answer is (3/11).

Question 9

The H.C.F. of $2^2 	imes 3^3 	imes 5^5$, $2^3 	imes 3^2 	imes 5^2 	imes 7$ and $2^4 	imes 3^4 	imes 5 	imes 7^2 	imes 11$ is

Accepted Answer

Concept & Formula: To find the Highest Common Factor (HCF) of numbers already given in their prime factorized form, you do not need to expand or multiply them out. The Rule for HCF of Prime Expressions: Select only the prime bases that are common to all the given numbers, and raise each to its lowest power present across the expressions. Step-by-Step Solution: • Given: We have three numbers expressed as products of primes: Number 1: 2^2 × 3^3 × 5^5 Number 2: 2^3 × 3^2 × 5^2 × 7 Number 3: 2^4 × 3^4 × 5 × 7^2 × 11 • Calculation / Deduction: 1. Identify Common Bases: Look at the prime bases (2, 3, 5, 7, 11). Only the bases 2, 3, and 5 appear in all three expressions. (We ignore 7 and 11 because they are missing from the first expression). 2. Find the Lowest Power of 2: The powers of 2 are 2^2, 2^3, 2^4. The lowest is 2^2. 3. Find the Lowest Power of 3: The powers of 3 are 3^3, 3^2, 3^4. The lowest is 3^2. 4. Find the Lowest Power of 5: The powers of 5 are 5^5, 5^2, 5^1 (since 5 is 5^1). The lowest is 5^1 or just 5. 5. Multiply these lowest powers together to get the HCF: HCF = 2^2 × 3^2 × 5 Exam Strategy & Shortcut: Rapid Elimination: You can solve this by scanning the options instantly. Since the first expression lacks 7 and 11, any option containing 7 or 11 is immediately incorrect. This quickly eliminates options (b) and (d). Next, look at the base 5. The third expression only has 5^1. Therefore, the HCF cannot have a power of 5 greater than 1. This eliminates option (c) which contains 5^5. Option (a) is the only survivor. Common Pitfall: Confusing the rules for HCF and LCM. Students often mistakenly select the *highest* powers or include all distinct prime factors (which calculates the LCM, not HCF). Remember: HCF equals Common bases only, Lowest powers. Final Answer: Therefore, the correct answer is 2^2 × 3^2 × 5.

Question 10

The H.C.F. of $2^4 	imes 3^2 	imes 5^3 	imes 7$, $2^3 	imes 3^3 	imes 5^2 	imes 7^2$ and $3 	imes 5 	imes 7 	imes 11$ is

Accepted Answer

Concept & Strategy: To find the Highest Common Factor (HCF) of numbers provided in their prime factorized form, you only need to look at the prime bases. The rule is: Identify the prime bases that are common to all expressions, and take the lowest power of each common base present. Multiply these together to get the HCF. Step-by-Step Solution: • Given: Expression 1: 2^4 × 3^2 × 5^3 × 7^1 Expression 2: 2^3 × 3^3 × 5^2 × 7^2 Expression 3: 3^1 × 5^1 × 7^1 × 11^1 • Calculation / Deduction: 1. Identify Common Bases: Look at the prime bases across all three expressions. The bases 3, 5, and 7 appear in every expression. The base 2 is missing from the third expression, and 11 is missing from the first two. 2. Lowest Power of 3: Comparing 3^2, 3^3, and 3^1, the lowest is 3^1 (or simply 3). 3. Lowest Power of 5: Comparing 5^3, 5^2, and 5^1, the lowest is 5^1 (or simply 5). 4. Lowest Power of 7: Comparing 7^1, 7^2, and 7^1, the lowest is 7^1 (or simply 7). 5. Calculate Final HCF: Multiply these common lowest powers together: HCF = 3 × 5 × 7 HCF = 15 × 7 = 105 Exam Strategy & Shortcut: Rapid Elimination: You can spot the answer in under 10 seconds. The third expression is just 3 × 5 × 7 × 11. Since 2 is missing, the HCF cannot be even. Options (c) 2310 and (d) 27720 are even numbers, so eliminate them instantly. Furthermore, 11 is not present in the first expression, so 11 cannot be in the HCF. Option (b) 1155 is divisible by 11 (1+5 = 6, 1+5 = 6). This leaves only 105 as the correct choice. Common Pitfall: The most frequent mistake is getting confused with the LCM rule and picking the *highest* powers, or blindly multiplying all distinct prime factors together. Always remember: HCF means "Common" bases and "Lowest" powers. Final Answer: Therefore, the correct answer is 105.

Question 11

H.C.F. of $4 	imes 27 	imes 3125$, $8 	imes 9 	imes 25 	imes 7$ & $16 	imes 81 	imes 5 	imes 11 	imes 49$ is

Accepted Answer

Concept & Strategy: Before applying the HCF rule for exponential expressions, all numbers must be converted into their prime factorized forms. The given numbers contain composite bases (like 4, 27, 25, 16, 81, etc.) which must be broken down first. Once converted to prime bases, the HCF is the product of the lowest powers of the common prime factors. Step-by-Step Solution: • Given: Expression 1: 4 × 27 × 3125 Expression 2: 8 × 9 × 25 × 7 Expression 3: 16 × 81 × 5 × 11 × 49 • Calculation / Deduction: 1. Convert Expression 1 to Primes: 4 = 2^2 27 = 3^3 3125 = 5^5 So, Exp 1 = 2^2 × 3^3 × 5^5 2. Convert Expression 2 to Primes: 8 = 2^3, 9 = 3^2, 25 = 5^2, 7 = 7^1 So, Exp 2 = 2^3 × 3^2 × 5^2 × 7^1 3. Convert Expression 3 to Primes: 16 = 2^4, 81 = 3^4, 5 = 5^1, 11 = 11^1, 49 = 7^2 So, Exp 3 = 2^4 × 3^4 × 5^1 × 11^1 × 7^2 4. Find Common Primes: The bases present in *all three* expressions are 2, 3, and 5. (Ignore 7 and 11 as they are not in Exp 1). 5. Extract Lowest Powers: Lowest power of 2 is 2^2 (from Exp 1) Lowest power of 3 is 3^2 (from Exp 2) Lowest power of 5 is 5^1 (from Exp 3) 6. Calculate HCF: HCF = 2^2 × 3^2 × 5^1 HCF = 4 × 9 × 5 = 180 Exam Strategy & Shortcut: Targeting the Smallest Component: Look for the smallest factor among the composite groupings. For base 5: The third expression has a standalone "5". This means the highest power of 5 in the HCF can only be 5^1 = 5. For base 2: The first expression has "4". The highest power of 2 in the HCF can only be 4. For base 3: The second expression has "9". The highest power of 3 in the HCF can only be 9. Multiply these limiting factors directly: 4 × 9 × 5 = 180. You don't need to calculate 3125 or 81 at all. Common Pitfall: Applying the HCF rule directly to the composite numbers without prime-factorizing them first. If you just look for "common numbers" in the raw expressions, you will completely miss the hidden prime factors shared across the composite bases. Final Answer: Therefore, the correct answer is 180.

Question 12

Find the highest common factor of $36$ and $84$.

Accepted Answer

Concept & Strategy: For relatively small numbers, the Highest Common Factor (HCF) can be found using the difference method. The HCF of two numbers will always evenly divide their difference. This narrows down the possibilities significantly. Alternatively, basic prime factorization or looking for the largest common divisor from the options works well. Step-by-Step Solution: • Given: The numbers are 36 and 84. • Calculation / Deduction: 1. Find the Difference: 84 - 36 = 48 2. The HCF must be a factor of 48. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. 3. Now, we check these factors against our given options, starting from the largest option to find the *highest* common factor. 4. Let's test Option (d) 18: 18 is not a factor of 48. (Skip) 5. Let's test Option (c) 12: 12 is a factor of 48. Let's see if it divides both 36 and 84: 36 \div 12 = 3 84 \div 12 = 7 6. Since 12 divides both perfectly, and we are working downwards from the largest options, 12 is the HCF. Exam Strategy & Shortcut: Option Testing from Largest to Smallest: Since you are looking for the *Highest* Common Factor, start testing the given options from largest to smallest. 1. Test 18: 36 is divisible by 18, but 84 \div 18 is not a whole number. 2. Test 12: 36 \div 12 = 3. 84 \div 12 = 7. Both divide cleanly. You have found your HCF. This bypasses all manual calculation and is incredibly fast for two-digit numbers. Common Pitfall: Starting the option testing from the smallest number. If you check 4 or 6 first, you will find they *do* divide both 36 and 84. If you stop there, you will select a common factor, but not the *highest* common factor. Always test in descending order. Final Answer: Therefore, the correct answer is 12.

Question 13

The H.C.F. of $204$, $1190$ and $1445$ is

Accepted Answer

Concept & Strategy: For finding the HCF of three large numbers, the continuous division method can be lengthy. The most efficient approach is the Difference Method. The HCF of any set of numbers will always evenly divide the difference between any two of those numbers. By finding the difference between the two numbers closest to each other, we drastically reduce the pool of possible factors. Step-by-Step Solution: • Given: The numbers are 204, 1190, and 1445. • Calculation / Deduction: 1. Identify the two numbers with the smallest gap between them. Here, they are 1190 and 1445. 2. Calculate their difference: 1445 - 1190 = 255 3. The HCF of the three numbers *must* be a factor of 255. 4. Let's look at the given options: 17, 18, 19, 21. 5. Since 255 ends in 5, it is an odd number. Therefore, it cannot have any even factors. This immediately eliminates Option (b) 18. 6. Let's test if 255 is divisible by Option (a) 17: 255 \div 17 = 15 Since 17 × 15 = 255, 17 is a valid factor of the difference. 7. To be absolutely sure, verify if 17 divides the smallest original number, 204: 204 \div 17 = 12 8. Since 17 divides both the difference of the large numbers and the standalone small number perfectly, it is our HCF. Exam Strategy & Shortcut: Unit Digit Viability Check: Take the smallest number, 204, and test the options using unit digits. We need an option that, when multiplied by some integer, ends in 4. • (b) 18: 8 × 3 = 24, 8 × 8 = 64. Possible, but 204/18 fails. • (c) 19: 9 × 6 = 54. 19 × 10 = 190, plus 19 is 209. Fails. • (d) 21: 1 × 4 = 4. 21 × 10 = 210, so 21 × 9 = 189. Fails. • (a) 17: 7 × 2 = 14. Let's try a multiplier ending in 2, like 12. 17 × 10 = 170. 17 × 2 = 34. 170 + 34 = 204. Perfect match. Common Pitfall: Attempting to apply prime factorization or the ladder division method to 1190 and 1445. These numbers have large, obscure prime factors (like 17) which are difficult to spot mentally, leading to a massive loss of time during an exam. Always use the difference method for large values. Final Answer: Therefore, the correct answer is 17.

Question 14

Which of the following is a pair of co-primes?

Accepted Answer

Concept & Logic: Two numbers are considered co-prime (or mutually prime) if they share no common factors other than 1. In other words, their Highest Common Factor (HCF) must be exactly 1: HCF(a, b) = 1 Step-by-Step Solution: Let's analyze the factors of each pair given in the options to find their HCF: • Option (a) (16, 62): Both numbers are even. Therefore, they are at least divisible by 2. The HCF is not 1. • Option (c) (21, 35): Both numbers are multiples of 7 (7 × 3 = 21 and 7 × 5 = 35). The HCF is 7. • Option (d) (23, 92): 23 is a prime number. If we multiply 23 × 4, we get 92. Thus, 23 is a factor of 92, making their HCF 23. • Option (b) (18, 25): The prime factorization of 18 is 2 × 3^2. The prime factorization of 25 is 5^2. They share absolutely no common prime factors. Their HCF is 1. Exam Strategy & Shortcut: Use divisibility rules to quickly eliminate options. Any pair of even numbers (like 16 and 62) can be instantly crossed out. Check for obvious multiples (21 and 35 are clearly in the 7 times table). This leaves (18, 25) as the obvious answer within seconds. Common Pitfall: Students often confuse "co-prime" with "prime numbers." Neither 18 nor 25 are prime numbers individually, but they are co-prime *to each other* because they share no common factors. Do not look for pairs where both numbers are prime; look for pairs with no common overlap. Final Answer: Therefore, the correct answer is (18, 25).

Question 15

The H.C.F. of $3556$ and $3444$ is

Accepted Answer

Concept & Strategy: To find the Highest Common Factor (HCF) of two large numbers quickly, use the Difference Method. The fundamental rule states that the HCF of two numbers divides their difference. Therefore, the HCF is either the difference itself or a factor of that difference. Step-by-Step Solution: • Given: We need to find the HCF of 3556 and 3444. • Step 1: Calculate the difference between the two numbers. 3556 - 3444 = 112 • Step 2: The HCF must be a factor of 112. Let's test the given options against the number 112. • 112 \div 23 \approx 4.86 (Not a factor) • 112 \div 25 = 4.48 (Not a factor) • 112 \div 26 \approx 4.30 (Not a factor) • 112 \div 28 = 4 (Exact factor!) • Step 3: Since 28 is the only option that cleanly divides the difference (112), it must be the HCF. (Both 3556 and 3444 are divisible by 28). Exam Strategy & Shortcut: Take the difference (112). Scan the options and look at the unit digits. To get a unit digit of 2 in 112, multiplying by a number ending in 8 (28 × 4) is the most logical step. This saves you from having to test divisibility for every single option. Common Pitfall: Using the continuous long division method for finding the HCF here will work, but it consumes valuable time during an exam. Relying on primary school methods instead of competitive aptitude shortcuts (like the difference rule) is the biggest reason students fail to finish quantitative sections on time. Final Answer: Therefore, the correct answer is 28.

Question 16

Find the lowest common multiple of $24$, $36$ and $40$.

Accepted Answer

Concept & Strategy: The Lowest Common Multiple (LCM) of a set of numbers is the smallest positive integer that is uniformly divisible by all the numbers in the set. The most structured way to find the LCM of multiple integers is to use Prime Factorization. Step-by-Step Solution: • Step 1: Prime factorize each number. • 24 = 8 × 3 = 2^3 × 3^1 • 36 = 4 × 9 = 2^2 × 3^2 • 40 = 8 × 5 = 2^3 × 5^1 • Step 2: Apply the LCM Rule. The LCM is found by taking the highest power of each unique prime factor present across all the numbers. • Highest power of 2 is 2^3 • Highest power of 3 is 3^2 • Highest power of 5 is 5^1 • Step 3: Calculate the final value. LCM = 2^3 × 3^2 × 5^1 LCM = 8 × 9 × 5 LCM = 72 × 5 LCM = 360 Exam Strategy & Shortcut: Use the Option Elimination Strategy. The LCM must be a multiple of 40, meaning it must end in 0 (all options do). It must be a multiple of 36, which means it must be divisible by 9 (sum of digits must be a multiple of 9). • 120: Sum = 3 (Not divisible by 9) • 240: Sum = 6 (Not divisible by 9) • 360: Sum = 9 (Divisible by 9!) • 480: Sum = 12 (Not divisible by 9) Only 360 satisfies the divisibility rule for 9, making it the correct answer in seconds without full calculations. Common Pitfall: Using the "ladder" or common division method can sometimes lead to arithmetic errors if you skip a prime factor. Furthermore, students often forget to check divisibility by 9 when 36 or 72 are in the mix, missing out on a massive time-saving shortcut. Final Answer: Therefore, the correct answer is 360.

Question 17

The L.C.M. of $22$, $54$, $108$, $135$ and $198$ is

Accepted Answer

Concept & Strategy: The Lowest Common Multiple (LCM) of a set of numbers is the smallest integer divisible by all numbers in the set. For larger numbers, use Prime Factorization. The LCM is the product of the highest power of each unique prime factor present. Step-by-Step Solution: • Step 1: Express each number as a product of its prime factors. • 22 = 2^1 × 11^1 • 54 = 2^1 × 3^3 • 108 = 2^2 × 3^3 • 135 = 3^3 × 5^1 • 198 = 2^1 × 3^2 × 11^1 • Step 2: Identify all unique prime bases: 2, 3, 5, and 11. • Step 3: Select the highest exponent for each prime base. • Highest power of 2 is 2^2 (4) • Highest power of 3 is 3^3 (27) • Highest power of 5 is 5^1 (5) • Highest power of 11 is 11^1 (11) • Step 4: Multiply these together. LCM = 2^2 × 3^3 × 5 × 11 LCM = 4 × 27 × 5 × 11 LCM = 108 × 55 = 5940 Exam Strategy & Shortcut: Use divisibility rules on the options. The LCM must be a multiple of 198, which is 18 × 11. Therefore, the answer must be divisible by 11. Apply the divisibility rule for 11 (difference between sum of alternating digits must be 0 or a multiple of 11): • (a) 330 \rightarrow (3+0) - 3 = 0 (Divisible by 11) • (b) 1980 \rightarrow (1+8) - (9+0) = 0 (Divisible by 11) • (c) 5940 \rightarrow (5+4) - (9+0) = 0 (Divisible by 11) • (d) 11880 \rightarrow (1+8+0) - (1+8) = 0 (Divisible by 11) Since all are divisible by 11, check divisibility by 27 (from 108). The sum of digits of 5940 is 18, which is divisible by 9, and 5940 \div 9 = 660, not divisible by 3. Wait, let's check 108. 5940 \div 108 = 55. Yes, it works. For speed, calculate the LCM(108, 135) = 540, then LCM(540, 198) = 5940. Common Pitfall: The continuous long division method is highly prone to arithmetic errors with a large set of numbers like this. Prime factorization is much safer and easier to track. Final Answer: Therefore, the correct answer is 5940.

Question 18

The H.C.F. of $\frac{2}{3}, \frac{8}{9}, \frac{64}{81}$ and $\frac{10}{27}$ is :

Accepted Answer

Concept & Formula: To find the Highest Common Factor (HCF) of a set of fractions, use the standard fraction formula: H.C.F. = \frac{H.C.F. of Numerators}{L.C.M. of Denominators} Step-by-Step Solution: • Given Fractions: (2/3), (8/9), (64/81), and (10/27). • Step 1: Find the HCF of the numerators (2, 8, 64, 10). The factors are 2, 2^3, 2^6, and 2 × 5. The highest common factor shared by all four numbers is clearly 2. • Step 2: Find the LCM of the denominators (3, 9, 81, 27). The denominators are all powers of 3: 3 = 3^1 9 = 3^2 27 = 3^3 81 = 3^4 The LCM of numbers with the same base is simply the one with the highest power, which is 81. • Step 3: Combine them into the final fraction. Final H.C.F. = (2/81) Exam Strategy & Shortcut: You can solve this visually in three seconds. Look at the numerators (2, 8, 64, 10); the smallest number is 2, and 2 divides all the others perfectly, so the top must be 2. Eliminate options (c) and (d). Look at the denominators; they are all multiples of 3, with 81 being the largest and a multiple of the others. So the bottom must be 81. Option (b) is the instant winner. Common Pitfall: Students who forget the formula often calculate the HCF for both numerators and denominators (resulting in (2/3)) or calculate the LCM for both (resulting in (160/81)). Always remember: "HCF on top for HCF questions." Final Answer: Therefore, the correct answer is (2/81).

Question 19

The H.C.F. of $\frac{9}{10}, \frac{12}{25}, \frac{18}{35}$ and $\frac{21}{40}$ is

Accepted Answer

Concept & Formula: When calculating the Highest Common Factor (HCF) of several fractions, we apply the foundational fraction rule: H.C.F. = \frac{H.C.F. of Numerators}{L.C.M. of Denominators} Step-by-Step Solution: • Given Fractions: (9/10), (12/25), (18/35), and (21/40). • Step 1: Calculate the HCF of the numerators (9, 12, 18, 21). All of these numbers are clearly multiples of 3. 9 = 3 × 3 12 = 3 × 4 18 = 3 × 6 21 = 3 × 7 The highest number that divides all of them is 3. So, the numerator of our answer is 3. • Step 2: Calculate the LCM of the denominators (10, 25, 35, 40). Prime factorize each: 10 = 2 × 5 25 = 5^2 35 = 5 × 7 40 = 2^3 × 5 Take the highest power of each prime factor (2, 5, 7): LCM = 2^3 × 5^2 × 7 LCM = 8 × 25 × 7 LCM = 200 × 7 = 1400 • Step 3: Assemble the fraction. Result = (3/1400) Exam Strategy & Shortcut: Calculate the HCF of the numerators first, which is 3. Instantly eliminate option (b) and (d). You are left with (3/5) and (3/1400). Now look at the denominators. For the denominator of the answer to be 5, the LCM of 10, 25, 35, 40 would have to be 5, which is logically impossible (the LCM must be greater than or equal to the largest number, 40). Thus, without doing the heavy math, option (c) is the only mathematically viable choice. Common Pitfall: Calculating the full LCM for the denominators using the ladder method wastes a lot of time. In multiple-choice exams, determining just the numerator and doing a quick logical check on the remaining denominator options is the key to finishing on time. Final Answer: Therefore, the correct answer is (3/1400).

Question 20

The L.C.M. of $\frac{1}{3}, \frac{5}{6}, \frac{2}{9}, \frac{4}{27}$ is

Accepted Answer

Concept & Formula: To find the Least Common Multiple (LCM) of a set of fractions, we use the standard formula: L.C.M. of fractions = \frac{L.C.M. of Numerators}{H.C.F. of Denominators} Step-by-Step Solution: • Given Fractions: (1/3), (5/6), (2/9), and (4/27). • Step 1: Calculate the L.C.M. of the numerators (1, 5, 2, 4). The multiples of the highest number (5) that are also divisible by 4 and 2 is 20. L.C.M. (1, 5, 2, 4) = 20 • Step 2: Calculate the H.C.F. of the denominators (3, 6, 9, 27). The factors are 3, 2 × 3, 3^2, and 3^3. The highest common factor shared by all is clearly 3. H.C.F. (3, 6, 9, 27) = 3 • Step 3: Assemble the fraction. Result = (20/3) Exam Strategy & Shortcut: First, scan the numerators: 1, 5, 2, 4. Their L.C.M. is obviously 20. Look at the options—only option (c) has 20 in the numerator. You can confidently select (c) without even calculating the H.C.F. for the denominators. Common Pitfall: Students often mix up the formula, calculating the L.C.M. of the denominators and the H.C.F. of the numerators (which would give (1/54), option a). Remember: "Whatever they ask for goes on top." If they ask for L.C.M., put L.C.M. in the numerator. Final Answer: Therefore, the correct answer is (20/3).

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